Centrifugal force
Centrifugal force is a force that is virtual but works similarly to the centripetal force. The centripetal force is the force acting towards the centre of a rotating system to keep the rotating object rotating. If you take the perspective of being in the rotating object, instead of the centre of the system as previously, then you are not rotating from your view; in fact the centre of the system is rotating about you. However there is still a centripetal force that should be acting on you to make you move towards the centre, since you are not moving from your point of view it appears that another force, equal and opposite, is acting on you to prevent the centripetal force moving you. This apparent force is the centrifugal force. Home Contents __TOC__ = The Death Star = The first death star is 160km in diameter1. In this we shall assume that the death star has no artificial gravity generators and relies on its rotation for centrifugal force to see if they could work in real life (even though in the films the millennium falcon lands perpendicular to the surface entrance and would fly out if this were the case). If we also assume that normal gravity for people within the star wars universe is similar to ours and that their tolerances are also similar then the majority of the station wants to be in a gravity range of ~0g-1.5g2 (Though please be aware that the 0g allowance is due to known long term stays on the ISS which although is safe leads to health problems later when returning to higher gravity; also the range of long term human gravity tolerance is not well know). Since 0g would not be useful for working on a station designed for people under gravity we will take the lower limit to be 0.5g (somewhat arbitrarily, although it was noted that it should be higher than moon gravity (~0.2g) since Apollo astronauts had heavy suits on and still bounced around). The centrifugal force felt at a point on the death star a distance r''' from its centre is given by: {\textbf{F} = m\omega^2 \textbf{r}} , hence the acceleration in g is: {\textbf{a}=\frac{\omega^2 \textbf{r}}{\textbf{g}}} . Notice that it is proportional to '''r hence the force decreases towards the centre. Therefore in our case we want 1.5g to be present at the surface of the death star, hence we solve for its rotational speed with r=80km. This yields an angular velocity of ~0.014rad/s. This is a very manageable and realistic speed. Given our minimum constraint on gravity tolerance the minimum radius that can be occupied is ~25km. This is fine for the death star as it is stated that only a habitable crust exists and most of the interior is machinery1 To see that this is manageable in the star wars universe we just need to calculate the energy needed to spin the death star up to this speed. To do this we can work out its rotational kinetic energy, similarly to how we work out its linear kinetic energy: E=I\omega^2 , where I is the objects moment of inertia, instead of mass, assuming the death star is a solid sphere (which excluding some room for corridors it effectively is) then we can substitute in for its moment of inertia: E = \frac{2}{5}MR^2\omega^2 . If we arbitrarily assume that the death star is made of steel then given the density of steel (7850kg/m3)3 then the energy required to reach this angular velocity is: 8.4YJ (8.4x1024J). This may seem like a massive amount of energy, which it is, however it is very achievable in the star wars universe by the death star. Since the death star has the ability to destroy entire planets by having the whole planet explode (to separate from cases where for example a planet is split in two, which still destroys it) then its laser must be able to overcome the gravitational binding energy of the planets. This is the minimum energy needed for a system to no longer be bound gravitationally, i.e. the energy at which everything within a system has sufficient energy to counteract gravity so that in net the systems gravity no longer affects it. Since the death star causes the planets to start moving apart too (explosion) then it must provide more than this energy. For the Earth, assuming uniform density (yields higher value) the binding energy is 2.242x1032J4. Given that there are larger planets than the Earth and the death star doesn't appear to have planet destroying limits it can easily produce orders of magnitude above this value of energy. This means it could very easily produce enough energy to spin itself up to speed. Further reading: Coriolis force One other important thing to consider when mimicking gravity through rotation is the Coriolis force. This is another virtual force and acts on any mass with velocity moving while under rotation within the system (e.g. when you walk on Earth's surface (though tiny)). This force acts radially outwards causing objects to appear to curve into the ground. The acceleration from this force is as follows: \left|\textbf{a}\right| = 2\omega\left|\textbf{v}\right| . If conventional bullets were used in star wars then this would produce major arcing problems, with a muzzle velocity of 1295m/s5 then as the bullet moved it would feel an additional 3.7g acting on it. Luckily weapons in star wars are slow moving energy weapons and within the short distances possible to fire over on the death star would be unnoticeable. The only other major thing to check is if people would notice this in everyday tasks. Taking the average human mass to be 65kg I believe a 10% (6.5kg equivalent) increase in weight would be noticeable for most people; therefore an Coriolis acceleration of 0.1g shall be considered the cut off point for acceptable. This leads to a maximum velocity for an object (person) moving within the death star to be 35m/s. This bound is more than sufficient for everyday activity. Given that the fastest a person may run is ~45km/h (12m/s)6 and things like fork lift trucks that may be operated move at ~30km/h the affect of this force shouldn't significantly impede everyday activities whilst aboard the death star. Concluding point Therefore in conclusion the death star could easily mimic gravity through rotation, throughout most of its internal structure (in fact more than needed in canon), given the readily available energy sources in star wars. Also due to the death star size the rotational velocity is so minimal, at 0.014rad/s (~7.5 minutes per rotation), that no arcing problems (from the Coriolis force) would be noticed in normal operation on board. References 1 -http://starwars.wikia.com/wiki/Death_Star, as of rogue one 2 -The Biology of Human Survival, Claude A. Piantadosi; https://www.quora.com/What-is-the-range-of-gravity-that-human-beings-can-tolerate-in-the-long-term 3 -http://www.engineeringtoolbox.com/metal-alloys-densities-d_50.html 4 -http://typnet.net/Essays/EarthBind.htm 5 -https://en.wikipedia.org/wiki/Muzzle_velocity 6 -https://en.wikipedia.org/wiki/Footspeed